Kevin Lowe's calculation is correct based on his assumptions. At the risk of lecturing a short course on statistics and probabilities follows:
To arrive at the probability of a series of events happening you must multiply them together - the more events the more items to multiply and the lower overall probability. Thus the odds of heads occurring two times in a row are 25% (0.5 x 0.5). (Accounting for all the possibilities: 25% - two heads, 25% - two tails, and 50% one each).
Now, let's put some real (unreal?) possibilities to Kevin's abstract example - Let's assign events that may have occurred to each variable:
X = Amanda bought/stole bleach on the morning of Nov 2nd.
Y = Amada and Raffaele were at the basketball court at 9:30 PM on Nov 1st
Z = Amanda carried a large knife from RS's kitchen in her bag for protection.
Further, Kevin conservatively assigned 60% probablility to each event actually happening - which is a generous hypothetical which I understood to be the probability equivalent of "more likely than not".
Thus the probability that ALL THREE happened is 0.6 x 0.6 x 0.6 = .213 or 21.3%. BTW - assuming 100% certitude for one event (which I don't) only increases the probability to 0.36 or 36%.
Machiavelli, reversed the calculation as follows: 0.4 x 0.4 x 0.4 = 0.064
AND he subratcted that result from 1.00 to get = 0.936 or 93.6%.
But that answers a different question than Kevin posed. Kevin is saying there is only a 21%+ chance that all three events occurred whereas Machiavelli calculates the chance that NONE of them occurred at all. I think he wants us to believe that such a high probability means no reasonable doubt.
I think it is obvious which calculation actually reveals probable doubt.
BTW - generally statistical certitude for probabilities is generally accepted to be 2 or 3 standard deviations. Given a traditional bell curve distribution 2 SD = 95% or 3 SD = 99%. Thus, Machiavelli's calculation even if you accept its logical underpinnings does not meet the statistical minimum of 5% (the inverse of 95%) let alone the 3 sd standard.
How do 2 SD and 3 SD apply to justice? A 2 SD system considers it acceptable that 5% of all trials convict innocent people. A 3 SD system accepts only 1%.
No, it's a fair deal more complicated than that to work out the statistical probabilities in this case. It's not simply a case of multiplying the individual probabilities. If it were, then you'd get the anomalous result that one piece of evidence that was 99% "proof of guilt" was massively better proof than 200 pieces of evidence that each were 99% "proof of guilt" - 0.99^200 = 13.4%.
Instead, this is a case of nested conditional probabilities. If we say that A = guilt, and the three pieces of evidence are x, y and z, then, statistically, what is being expressed is the probability of A
given x - P(A given x), which is (arbitrarily) given as 0.6 in this instance. And, similarly, P(A given y) and P(A given z) are both also 0.6.
The question then becomes, if P(A given x), P(A given y) and P(A given z) each equal 0.6, then what is P(A given x,y,z)?
This is a complex statistical puzzle. And it's most definitely not answered by multiplying together the three original conditional probabilities. So I'd be extremely interested to see Machiavelli's working behind his 93.6% answer.
ETA I just realised that you'd provided Machiavelli's calculations for him within this post. And he's performing a thoroughly bogus calculation to reach his answer as well (if indeed this is the way he calculated his answer). But I'd still be interested in hearing more of his statistical acumen....